3) Show that the roots of the equation
(x-a)(x+b) + (x-b) (x-c) + (x-c)(x-a) = 0
are always real and they cannot be equal unless a = b = c.
Answers
I think question is like this
(x - a)(x - b)+(x - b)(x - c )+(x- c )(x - a) =0
x² -ax - bx +ab +x² -cx - bx +bc +x² -ax +cx +ac =0
3x² - ( 2a + 2b + 2c )x + (ab + bc + ca )=0
It is a qyadratic equation comparing it with Ax² + Bx + C = 0 we get
A = 3 , B = - (2a+2b+2c) , C = (ab+bc+ca)
Now we find value of discriminant
B² - 4AC
Putting values of A, B and C
={-(2a+2b+2c)}² - 4(3)(ab+bc+ca)
= 4a²+4b²+4c²+8ab+8bc+8ca-12ab-12bc- 12ca
= 4a²+4b²+4c²-4ab-4bc-4ca
=2( 2a² + 2b² + 2c² - 2ab - 2bc - 2ca )
= 2{(a²+b²-2ab)+(b²+c²-2bc)+(c²+a²-2ca)}
=2{( a- b)² + ( b - c )² + ( c - a )²} ≥ 0
Because (a - b )² , (b - c )² and (c - a )² are alwas positive so discriminant is always positive So roots of given quadratic equation is always real
Roots are equal when discriminant is equal to zero it is possible only when
(a - b )² = 0 , ( b - c )² = 0 , ( c - a )² = 0
Because sum of squares is zero only when each of them is equal to zero
If (a - b )² = 0
a - b = 0
a = b
Similarly from other two relations we get
b = c and c = a
So a = b = c
So roots are equal if a = b = c